A= 1/3+1/6+1/10+1/15=....+1/45
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
CT
1
AH
Akai Haruma
Giáo viên
9 tháng 11 2021
Lời giải:
$\frac{A}{2}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$
$=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}+\frac{9-8}{8\times 9}+\frac{10-9}{9\times 10}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}$
$=1-\frac{1}{9}=\frac{8}{9}$
$\Rightarrow A=2\times \frac{8}{9}=\frac{16}{9}$
BN
1
H9
HT.Phong (9A5)
CTVHS
16 tháng 10 2023
\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\)
\(A=2\times\dfrac{1}{2}\times\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}\right)\)
\(A=2\times\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\right)\)
\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)
\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{11}\right)\)
\(A=2\times\dfrac{9}{22}\)
\(A=\dfrac{9}{11}\)
L
3
HT
10 tháng 9 2016
\(a=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(a=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+\frac{1}{3.5}+...+\frac{1}{5.9}\)
\(a=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(a=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
=> \(a=2.\frac{2}{5}\)
=> \(a=\frac{4}{5}\)
10 tháng 9 2016
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{45}\)
\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right)\cdot\frac{1}{2}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
VB
1
KN
1
S
24 tháng 6 2019
\(=1-2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{90}\right)=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\right)=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......-\frac{1}{10}\right)=1-2\left(\frac{1}{2}-\frac{1}{10}\right)=1-\frac{2.4}{10}=1-\frac{4}{5}=\frac{1}{5}\)
DD
Đoàn Đức Hà
Giáo viên
30 tháng 8 2021
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\right)\)
\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{4}{5}\)
4 tháng 5 2018
\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{36}\) + \(\dfrac{1}{45}\)
= \(\dfrac{2}{4}\) + \(\dfrac{2}{6}\) + \(\dfrac{2}{12}\) + \(\dfrac{2}{20}\) + \(\dfrac{2}{30}\) + ... + \(\dfrac{2}{72}\) + \(\dfrac{2}{90}\)
= \(\dfrac{2}{2.2}\) + \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + \(\dfrac{2}{5.6}\) + ... + \(\dfrac{2}{8.9}\) + \(\dfrac{2}{9.10}\)
= 2 (\(\dfrac{1}{2.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + ... + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\))
= 2 (\(\dfrac{1}{2}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{8}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\)) = 2 (\(\dfrac{1}{2}\) - \(\dfrac{1}{10}\)) = 2 . \(\dfrac{2}{5}\) = \(\dfrac{4}{5}\)
Mặc dù có 1 thiếu!
Nhân cả tử cả mẫu của các phân số trong A với 2 ta có:
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..........+\frac{2}{90}\)
\(=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.........+\frac{1}{90}\right)\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{9.10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=2.\frac{2}{5}\)
\(=\frac{4}{5}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{2}{5}\)